H E S are three numbers HE*HE=SHE,then S+H+E is
Answers
Answer:
The alphabets E, S, H can take any value from 0-9
\LARGE{\mathsf (HE)^H = SHE}(HE)
H
=SHE
\begin{gathered}\\\end{gathered}
\underline{\Huge{\textbf{Step-1:}}}
Step-1:
\textsf{Note the Range of values that RHS can take}Note the Range of values that RHS can take
From this it can be concluded that,
\begin{itemize}\begin{enumerate}\item \textsf{RHS is a 3-digit No. }\item \textsf{None of the digits in the RHS are same}\end{enumerate}\end{itemize}
\implies \sf SHE \in (111, 999)⟹SHE∈(111,999) ———[1]
\begin{gathered}\\\end{gathered}
\underline{\Huge{\textbf{Step-2:}}}
Step-2:
\textsf{Find the Range of values that HE(in LHS) can take}Find the Range of values that HE(in LHS) can take
\quad\maltese\;✠ Consider the least possible value for E and using \textsf{trial and Error Method}trial and Error Method to find the Range of values within which LHS falls so that [1] gets satisfied.
Here, Least possible value for an alphabet = 0
(Given that an alphabet can take any value from 0-9)
\therefore Take\: E = 0∴TakeE=0
\begin{itemize}\item Say H = 1\begin{itemize}\item $\rm HE^H \implies 10^1$=10\begin{itemize}\item $10 \notin (111, 999)$\item $\therefore\rm H \neq 1$\end{itemize}\end{itemize}\end{itemize}
\begin{itemize}\item Say H = 2\begin{itemize}\item $\rm HE^H \implies 20^2$ = 400\begin{itemize}\item $400 \in (111, 999)$\item $\therefore\rm H = 2$\end{itemize}\end{itemize}\end{itemize}
\begin{itemize}\item For H $ > $ 2\begin{itemize}\item $\sf HE \notin (111, 999)$\item $\because\texttt{RHS has to be a 3-digit No. }$\end{itemize}\end{itemize}
Clearly\: HE \in (20, 29]ClearlyHE∈(20,29]
\implies H= 2⟹H=2 ———[2]
\quad In the interval, 20 was excluded due to the fact that the Result obtained for (20)^2(20)
2
has two similar digits which doesn't satisfy the RHS (which has different values.)
\begin{gathered}\\\end{gathered}
\underline{\Huge{\textbf{Step-3:}}}
Step-3:
\textsf{Find the values that E(in LHS) can take}Find the values that E(in LHS) can take
\mathsf{HE^H =\underbrace{HE\timesHE \times HE\times HE \cdots\cdots HE}_{H\: times}}HE
H
=
Htimes
HE\timesHE×HE×HE⋯⋯HE
Consider the terms contributing in LHS to Result the Unit digit E.
(H\underline{E})^H = SH\underline{E}(H
E
)
H
=SH
E
\underbrace{E\timesE \times E\times E \cdots\cdots E}_{H\: times}\longrightarrow\textsf{Resulted Unit digit E}
Htimes
E\timesE×E×E⋯⋯E
⟶Resulted Unit digit E
\quad\maltese\;✠ It is observed that,
\begin{gathered}\textbf{E is number which when multiplied} \\\textbf{ by itself a certain number of times results} \\ \textbf{the same number at it's unit's place. }\end{gathered}
E is number which when multiplied
by itself a certain number of times results
the same number at it’s unit’s place.
If the No. has Unit digit 1, 5, 6, then any power of that No. will have the same unit digit.
In other words,
\begin{gathered}\boxed{\begin{minipage}{5cm}\maltese \; \; \textsf{Let $xy^z = k$,\\\\ If \; $y$} =\begin{cases}1 &\textsf{Unit digit in k is 1}\\5 &\textsf{Unit digit in k is 5}\\6 &\textsf{Unit digit in k is 6}\end{cases}\; \; \end{minipage}}\end{gathered}
\therefore E= 1\: or\: 5\: or\ 6∴E=1or5or 6 ———[3]
\begin{gathered}\\\end{gathered}
\underline{\Huge{\textbf{Step-4:}}}
Step-4:
\begin{gathered}\textsf{Check the possibilities by combining}\\\textsf{the results of [2] and [3] and note the result}\end{gathered}
Check the possibilities by combining
the results of [2] and [3] and note the result
Combining [2] & [3],
We have
\begin{gathered}\bf HE = \begin{cases}21\\25\\26\end{cases}\end{gathered}
HE=
⎩
⎪
⎪
⎨
⎪
⎪
⎧
21
25
26
\underline{(HE)}^H= S\underline{HE}
(HE)
H
=S
HE
\quad\maltese\;✠ It is also observed that,
\begin{gathered}\textbf{the tens Digit and Unit digit in}\\ \textbf{base of LHS is same as the tens Digit}\\ \textbf{and unit's digits in result of product.}\end{gathered}
the tens Digit and Unit digit in
base of LHS is same as the tens Digit
and unit’s digits in result of product.
21 doesn't satisfy this condition \sf(21^2= 441)(21
2
=441)
26 doesn't satisfy this condition \sf(22^2 = 484)(22
2
=484)
\begin{gathered}\textsf{The only Number that satisfies this}\\\textsf{condition is \Large{\textbf{25.}}}\end{gathered}
The only Number that satisfies this
condition is 25.
\begin{gathered}\\\end{gathered}
\boxed{\huge{\boxed{\LARGE{\mathsf ( \underline{25})^2 = 6 \underline{25}}}}}
(
25
)
2
=6
25
\therefore \begin{Large}\begin{array}{ccc}\bf S = 6;&\bf H = 2 ;& \bf E= 5\end{array}\end{Large}
\begin{gathered}\\\end{gathered}
\underline{\Huge{\textbf{Step-5:}}}
Step-5:
\textsf{Find S+H+E}Find S+H+E
\underline{\textsf{S+H+E =}}
S+H+E =
\underline{\textsf{6+2+5 = 13}}
6+2+5 = 13
\begin{gathered}\\\end{gathered}
\blacksquare\; \; \textsf{The value of S+H+E = \Huge{\underline{\LARGE{\textbf{13}}}}}■The value of S+H+E =
13